Cho a,b,c>0.Tìm Max:
\(A=\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\)
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G
23 tháng 5 2019
\(A=\frac{a\sqrt{a}}{\sqrt{a+b+2c}}+\frac{b\sqrt{b}}{\sqrt{b+c+2a}}+\frac{c\sqrt{c}}{\sqrt{c+a+2b}}\)
\(A=\frac{a^2}{\sqrt{a\left(a+b+2c\right)}}+\frac{b^2}{\sqrt{b\left(b+c+2a\right)}}+\frac{c^2}{\sqrt{c\left(c+a+2b\right)}}\)
\(\ge\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\)
Xét: \(2\left(\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\right)\)
\(=\sqrt{4a\left(a+b+2c\right)}+\sqrt{4b\left(b+c+2a\right)}+\sqrt{4c\left(c+a+2b\right)}\)
\(\le\frac{4a+a+b+2c+4b+b+c+2a+4c+c+a+2b}{2}=4\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\le2\left(a+b+c\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
25 tháng 1 2018
ÁP DỤNG BĐT COSI TA CÓ :\(\sqrt{\frac{a}{b+c+2a}}\le\frac{a}{b+c+2a}+\frac{1}{4}\)
\(\sqrt[]{\frac{b}{a+c+2b}}\le\frac{b}{a+c+2b}+\frac{1}{4}\)
\(\sqrt[]{\frac{c}{a+b+2c}}\le\frac{c}{a+b+2c}+\frac{1}{4}\)
ĐẶT A=\(\sqrt[]{\frac{a}{b+c+2a}}+\sqrt[]{\frac{b}{a+c+2b}}+\sqrt[]{\frac{c}{a+b+2c}}\)
\(\le\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}+\frac{3}{4}\)
ÁP DỤNG BĐT :\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Rightarrow\frac{a}{b+c+2a}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
\(\Rightarrow\frac{b}{a+c+2b}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(\Rightarrow\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{c+b}\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{1}{4}\left(1+1+1\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{3}{2}\)
DẤU = XẢY RA\(\Leftrightarrow a=b=c\)
30 tháng 8 2020
Một lời giải khác:
\(\left(\Sigma\sqrt{\frac{a}{b+c+2a}}\right)^2=\left(\Sigma\sqrt{\frac{a\left(a+2c+b\right)}{\left(a+2c+b\right)\left(b+c+2a\right)}}\right)^2\)
\(\le\left[\Sigma a\left(a+2c+b\right)\right]\left[\Sigma\frac{1}{\left(a+2c+b\right)\left(b+c+2a\right)}\right]=\Sigma\frac{a^2+3ab}{\left(a+2c+b\right)\left(b+c+2a\right)}\)
\(=\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\)
Cần chứng minh \(\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\le\frac{9}{4}\)
Chịu khó quy đồng :V
AH
Akai Haruma
Giáo viên
1 tháng 2 2020
Lời giải:
Với $a,b,c>0$ dễ thấy $0< \frac{a}{a+2b}< 1$
$\Rightarrow 0< \sqrt{\frac{a}{a+2b}}< 1$
$\Rightarrow \sqrt{\frac{a}{a+2b}}> \frac{a}{a+2b}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
$\text{VT}> \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}$
Áp dụng BĐT Cauchy-Schwarz:
$\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq \frac{(a+b+c)^2}{a^2+2ba+b^2+2cb+c^2+2ac}=1$
Do đó $\text{VT}>1$ (đpcm)
T
2 tháng 2 2020
Sử dụng BĐT AM-GM:
\(VT=\sum\limits_{cyc} \sqrt{\frac{a}{a+2b}} =\sum\limits_{cyc} \frac{a}{\sqrt{a(a+2b}}\geq \sum\limits_{cyc} \frac{2a}{2(a+b)}\)
\(=\sum\limits_{cyc} \frac{a^2}{a^2 +ab} \ge \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} >\frac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2bc+2ca} = 1\) (đpcm)
P/s: Em không chắc lắm.
TT
20 tháng 2 2020
Theo e nghĩ là đề phải như này cơ ạ :
\(\frac{a}{\sqrt{b+c+2a}}+\frac{b}{\sqrt{c+a+2b}}+\frac{c}{\sqrt{a+b+2c}}\le\frac{3}{2}\)
Biến đổi và sử dụng Cô - si là sẽ ra :
Ta có : \(\frac{a}{\sqrt{b+c+2a}}+\frac{b}{\sqrt{c+a+2b}}+\frac{c}{\sqrt{a+b+2c}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)+\left(a+c\right)}}+\frac{b}{\sqrt{\left(c+b\right)+\left(a+b\right)}}+\frac{c}{\sqrt{\left(a+c\right)+\left(b+c\right)}}\)
\(=\sqrt{\frac{a.a}{\left(a+b\right)+\left(a+c\right)}}+\sqrt{\frac{b.b}{\left(b+a\right)+\left(b+c\right)}}+\sqrt{\frac{c.c}{\left(c+a\right)+\left(c+b\right)}}\)
\(\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{c+b}\right)=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
T
20 tháng 2 2020
Đề không sai đâu:P
\(VT=\Sigma_{cyc}2\sqrt{\frac{1}{4}.\frac{a}{b+c+2a}}\le\Sigma_{cyc}\left[\frac{1}{4}+\frac{a}{\left(a+b\right)+\left(a+c\right)}\right]\)
\(\le\Sigma_{cyc}\left[\frac{1}{4}+\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}\right]=\frac{3}{2}\)
Trước hết ta chứng minh BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Thật vậy:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}.3\sqrt[3]{a.b.c}.3\sqrt[3]{ab.bc.ca}\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Ta có:
\(A^2=\left(\sqrt{a+c}.\sqrt{\frac{2a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{a+b}.\sqrt{\frac{2b}{\left(a+b\right)\left(b+c\right)}}+\sqrt{b+c}\sqrt{\frac{2c}{\left(c+a\right)\left(b+c\right)}}\right)^2\)
\(\Rightarrow A^2\le\left(a+c+a+b+b+c\right)\left(\frac{2a}{\left(a+b\right)\left(a+c\right)}+\frac{2b}{\left(a+b\right)\left(b+c\right)}+\frac{2c}{\left(c+a\right)\left(b+c\right)}\right)\)
\(\Rightarrow A^2\le\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)}=9\)
\(\Rightarrow A\le3\)
\(A_{max}=3\) khi \(a=b=c\)